$\overline{AC}$ is $9$ units long $\overline{BC}$ is $3$ units long $\overline{AB}$ is $3\sqrt{10}$ units long What is $\sin(\angle BAC)$ ? $A$ $C$ $B$ $9$ $3$ $3\sqrt{10}$
Answer: SOH CAH TOA in = pposite over ypotenuse opposite $= \overline{BC} = 3$ hypotenuse $= \overline{AB} = 3\sqrt{10}$ $\sin(\angle BAC)=\frac{3}{3\sqrt{10}}$ $=\dfrac{ \sqrt{10}}{10}$